1
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 9th January Evening Slot

A power transmission line feeds input power at 2300 V to a srep down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :
A
50 A
B
45 A
C
35 A
D
25 A

## Explanation

Given,

Primary voltage (VP) = 2300 V

Primary current (IP) = 5A

Secondary voltage (VS) = 230 V

efficiency ($\eta$) = 90%

We know,

Efficiency ($\eta$) = ${{{\mathop{\rm Sec}\nolimits} ondary\,Power} \over {\Pr imary\,\,Power}}$

$\therefore$   $\eta$ = ${{{P_S}} \over {{P_P}}}$ = ${{{V_S}\,{I_S}} \over {{V_P}\,{I_P}}}$

$\Rightarrow$   0.9 = ${{230 \times {{\rm I}_S}} \over {2300 \times 5}}$

$\Rightarrow$   I = 45 A
2
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 10th January Morning Slot

If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6 sin $\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$ then the maximum electric field associated with it is -
A
4.5 $\times$ 104 N/C
B
4 $\times$ 104 N/C
C
6 $\times$ 104 N/C
D
3 $\times$ 104 N/C

## Explanation

E0 = B0 $\times$ C

= 100 $\times$ 10$-$6 $\times$ 3 $\times$ 108

= 3 $\times$ 104 N/C

$\therefore$  correct answer is 3 $\times$ 104 N/C
3
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 10th January Evening Slot

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -
A
740 J
B
637.5 J
C
540 J
D
437.5 J

## Explanation

$L{{di} \over {dt}} = 25$

$L \times {{15} \over 1} = 25$

$L = {5 \over 3}H$

$\Delta E = {1 \over 2} \times {5 \over 3} \times ({25^2} - {10^2})$

$= {5 \over 6} \times 525 = 437.5$ J
4
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 10th January Evening Slot

The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$ The magnetic field $\overrightarrow B$(x,z, t) is given by $-$ (c is the velocity of light)
A
${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$
B
${1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$
C
${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$
D
${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$

## Explanation

$\overrightarrow E = 10\widehat j\cos \left[ {\left( {6\widehat i + 8\widehat k} \right).\left( {x\widehat i + z\widehat k} \right)} \right]$

$= 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$

$\therefore$   $\overrightarrow K = 6\widehat i + 8\widehat k;$ direction of waves travel.

i.e., direction of 'c'.

$\therefore$   Direction of $\widehat B$ will be along

$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$

Mag. of $\overrightarrow B$ will be along

$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$

Mag. of $\overrightarrow B = {E \over C} = {{10} \over C}$

$\therefore$   $\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}$

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